A sum fetched a total simple interest of Rs. 4016.25 at the rate of 9% p.a. in 5 years. What is the sum?
Answer: D
\[ \text{SI} = \frac{P \times R \times T}{100} \]
\[ 4016.25 = \frac{P \times 9 \times 5}{100} = 0.45P \]
\[ P = \frac{4016.25}{0.45} = \text{Rs. } 8925 \]
Five years at 9% comes to 45% of the sum, so the principal is a little over twice the interest.
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What sum of money will produce Rs.70 as simple interest in 4 years at 3 1/2 percent?
Answer: B
\[ \text{SI} = \frac{P \times R \times T}{100} \]
\[ 70 = \frac{P \times 3.5 \times 4}{100} = 0.14P \]
\[ P = \frac{70}{0.14} = \text{Rs. } 500 \]
Over four years at 3.5% the interest comes to 14% of the sum.
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4000 was divided into two parts such a way that when first part was invested at 3% and the second at 5%, the whole annual interest from both the investments is Rs.144, how much was put at 3%?
Answer: C
Let \( p \) be the part at 3%; the rest, \( 4000 - p \), earns 5%.
\[ 0.03p + 0.05(4000 - p) = 144 \]
\[ 0.03p + 200 - 0.05p = 144 \quad\Longrightarrow\quad -0.02p = -56 \]
\[ p = \text{Rs. } 2800 \]
Checking: Rs. 84 from the 3% part and Rs. 60 from the Rs. 1200 at 5%.
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At what rate percent on simple interest will Rs.750 amount to Rs.900 in 5 years?
Answer: C
\[ \text{SI} = \frac{P \times R \times T}{100} \]
\[ \text{SI} = 900 - 750 = \text{Rs. } 150 \]
\[ 150 = \frac{750 \times R \times 5}{100} = 37.5R \]
\[ R = 4\% \]
The Rs. 900 is the amount, so the principal must be subtracted first.
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Manoj borrowed Rs. 3450 from Anwar at 6% p.a. simple interest for three years. He then added some more money to the borrowed sum and lent the whole of it to Ramu for the same time at 9% p.a. simple interest. If Manoj gains Rs. 824.85 by way of interest from the whole transaction, what is the sum he lent to Ramu?
Answer: B
\[ \text{SI} = \frac{P \times R \times T}{100} \]
What Manoj pays Anwar on the borrowed Rs. 3450 at 6% for 3 years:
\[ \frac{3450 \times 6 \times 3}{100} = \text{Rs. } 621 \]
What he receives from Ramu on the whole sum \( S \) at 9% for 3 years:
\[ \frac{S \times 9 \times 3}{100} = 0.27S \]
His gain is the difference:
\[ 0.27S - 621 = 824.85 \quad\Longrightarrow\quad 0.27S = 1445.85 \]
\[ S = \text{Rs. } 5355 \]
So he added \( 5355 - 3450 = \text{Rs. } 1905 \) of his own money.
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A person lent a certain sum of money at 4% per annum at simple interest and in 8 years the interest amounted to Rs.340 less than the sum lent. What was the sum lent?
Answer: D
\[ \text{SI} = \frac{P \times R \times T}{100} \]
The interest falls Rs. 340 short of the principal.
\[ \frac{P \times 4 \times 8}{100} = P - 340 \]
\[ 0.32P = P - 340 \quad\Longrightarrow\quad 0.68P = 340 \]
\[ P = \text{Rs. } 500 \]
Checking: the interest is Rs. 160, which is indeed Rs. 340 less than Rs. 500.
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Raman paid Rs. 11400 as interest after 9 years. He had borrowed some money at rate of 6% for first two years, 9% for next three years and 14% for rest of the period. How much money did he borrow?
Answer: C
\[ \text{SI} = \frac{P \times R \times T}{100} \]
The rate changes with the period, so add the three slabs. Beyond five years leaves \( 9 - 5 = 4 \) years at 14%.
\[ P\left(\frac{6 \times 2}{100} + \frac{9 \times 3}{100} + \frac{14 \times 4}{100}\right) = 11400 \]
\[ P(0.12 + 0.27 + 0.56) = 0.95P = 11400 \]
\[ P = \text{Rs. } 12{,}000 \]
The final slab runs for four years, not nine - "beyond five years" means the remainder of the nine.
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A certain sum amounts to Rs.1725 in 3 years and Rs.1875 in 5 years. Find the rate % per annum?
Answer: B
\[ \text{SI} = \frac{P \times R \times T}{100} \]
Under simple interest the yearly addition is constant, so the gap between the two amounts covers two years.
\[ 1875 - 1725 = 150 \text{ over 2 years} \quad\Longrightarrow\quad 75 \text{ per year} \]
\[ P = 1725 - 3 \times 75 = \text{Rs. } 1500 \]
\[ R = \frac{75 \times 100}{1500} = 5\% \]
Working back three years from the first amount recovers the principal.
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Find the principle on a certain sum of money at 5% per annum for 2 2/5 years if the amount being Rs.1120?
Answer: A
\[ \text{SI} = \frac{P \times R \times T}{100} \]
The period \( 2\tfrac{2}{5} \) years is 2.4 years, and Rs. 1120 is the amount.
\[ 1120 = P\left(1 + \frac{5 \times 2.4}{100}\right) = P(1.12) \]
\[ P = \text{Rs. } 1000 \]
The interest is Rs. 120, which is 12% of the principal - 5% a year for 2.4 years.
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A sum of money lent out at S.I. amounts to Rs. 720 after 2 years and to Rs. 1020 after a further period of 5 years. The sum is?
Answer: B
\[ \text{SI} = \frac{P \times R \times T}{100} \]
The five years between the two figures account for the whole increase.
\[ \frac{1020 - 720}{5} = \frac{300}{5} = \text{Rs. } 60 \text{ per year} \]
\[ P = 720 - 2 \times 60 = \text{Rs. } 600 \]
"A further period of 5 years" means seven years in all, not five - but only the gap of five is needed.
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